Proof that $M(x)$ is little o of x?

Question

Let $M(x)$ be Mertens' function. Seeking contradiction, assume that $M(x)$ is not $o(x)$. Then, since $M(x) = O(x)$, putting this statement together with the negation of $o(x)$, through the inequality definitions of both notations, one can get that $M(x) = \Theta(x)$. Using Abel's lemma: $$\sum_{n\le x} \frac{\mu(n)}{n} = \frac{M(x)}{x} + \int_1^x \frac{M(t)}{t^2}dt$$ Replacing $M(x)$ by $\Theta(x)$ on the integral we get $$O(1) + \int_1^x \Theta(\frac{1}{t})dt = \Theta(log x)$$ So $\sum_{n\le x} \frac{\mu(n)}{n}$ grows like $log (x)$ and therefore is unbounded, but using dirichlets hyperbola method, one can get that $$1 = x\sum_{n\le x} \frac{\mu(n)}{n} + O(x)$$ Which means the sum is actually bounded, so we get a contradiction implied by our original assumption that $M(x)$ is not $o(x)$, therefore it must be $o(x)$. Does this proof work or is there anything in it that fails?

Answer 1

In general $$ f\in\Theta(g)\not\to \sum_{k=1}^nf(k)\in \Theta\left(\sum_{k=1}^ng(k)\right),$$ and even less does the similar claim with integrals hold.

As a counterexample (even with $f$ non-negative), consider $$ f(n)=\begin{cases}n,&n\text{ is a power of }2\\0,&\text{otherwise}\end{cases},$$ and $ g(n)=2n-1$ (so $\sum g(k)=n^2$). Then clearly $f\in O(g)$ and $f\notin o(g)$. But for $2^m\le n<2^{m+1}$, we have $$\sum_{k=1}^nf(n)=\sum_{j=0}^m2^j=2^{m+1}-1<2n,$$ so that $$\sum_{k=1}^nf(n)\in o\left(\sum_{k=1}^ng(n)\right).$$ Therefore, your computations do not allow the conclusion that $\sum \frac{\mu(n)}n$ is unbounded.